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Special Relativity:
Michelson, Albert (18521931)
(image from
scienceworld.wolfram.com)
In 1887 Michaelson and Morely used their
interferometer to search for the aether.
Aether Wind
V = 30 km/s
V = 30 km/s
E
To distant star
Aether Wind
V = 30 km/s
sun
To distant star
Wavefront from
distant star
This apparatus would in principle, detect the
motion of the Earth about the sun by
studying the speed of light from a distant
star as the Earth went towards it and
perpendicular to it as it orbited.
(image from scienceworld.wolfram.com)
Eath's speed around the sun is: v = 2R/T =
29 km/s.
While the Earth is in orbit, the relative speed
of light towards and away from a star was
thought to be:
V = c + v where v equals 29 km/s, and c =
300,000 km/s. MM felt they could detect
this difference.
The Basic Idea:
by analizing the time of travel for the two
beams (1) and (2), we can determine the
time difference at the telescope
t  t1  t2
And the corresponding phase difference:

2
t
T
For the “parallel” Beam:
Vearth
c–v
c+v
For the beam parallel to the motion of the
earth, the speed of light would be c +v going
“downstream” and c-v going upsrream.
Since time equals speed/distance, we can see
that
L1
L1
t1 

cv cv
2L1c
2L1c
t1  2 2  2
c  v c (1 v 2
t1 
2L1
v2
c(1 2 )
c
c
2
)
Before we go on…Check out this applet:
http://www.phy.ntnu.edu.tw/java/relativity/relativity.html
For the Transverse component:
c(t2/2)
Ve(t2/2)
t2 2
(ct2 )  (v )  ( L2 )2
2
2
L2
after a "bit" of algebra this becomes:
2L2
t2 
c (1 v2 c2 )
so...t 
2
L1
L2
 (
)
2
2 
2
2
c (1 v c )
(1 v c )
Equation I
if ...L1  L2  L
2L
1
1
t 
(

)
2
2
2
2
c (1 v c )
(1 v c )
Now since v/c << 1, we can use the
binomial theorem:
(1 + x)n = 1 + nx so that
2L
((1  v 2 c 2 )  (1 v 2 2c 2 )) 
c
L v2
 ( 2)
c c
t 
now let us rotate the apparatus by 90
degrees, or let the Earth travel 90 degrees
around the sun, so that L1 is now
perpendicular to the “aether wind”. In this
case, we now have a different time shift t’
2L1
t perp 
c (1 v2 c 2 )
t parallel
2L2

c(1 v2 c 2 )
so...t ' 
2
2L1
2L2
 (

)
2
2
2
2
c c (1 v c ) c(1 v c )
if ...L1  L2  L, and
using the binomial theorm
as before,
2
L
v
t '   ( 2 )
c c
note that a " time shift"
is just like a path length
difference
t 

T 2
where  is the phase difference :
i.e.  = (2m +1)
for destructive interference
So rotating the apparatus,
or waiting three months
should produce a changing
Fringe pattern :
2
(t - t ' )     '  2m
T
substituting T =

c
and
L v2
t - t  2 ( 2 ) gives :
c c
2 2L v2
( ( 2 ))  2m
 c c
c
or...
'
2L v2
( 2 )  m = fringe shift.
 c
Using their longest interferometer
L = 11m,  = 500 nm, gives :
2(11)
30 km s
2
m =
(
)
5x10 -7 300,000 km s
m = .44 fringes
so size (L) is very important...
but they could see a shift this
small by changing the length of
one arm, but no matter
how long they looked as
they earth orbited the sun,
they saw.....
no fringe shift!
The Lorentz Solution
the lack of a fringe shift didn' t
immediately kill the aether
hypothysis
Conrad Lorentz proposed
that the arm of the MM apparatus
Parallel to the Aether wind
contracted by an amount :
L parallel  L perp
2
v
1
c2
substituting this result in
equation 1 gives :
2
2
Lperp
2 L perp (1 v c )
 (

)
2
2
2
2
c
(1 v c )
(1 v c )
L perp
L perp
2
 (

)
2
2
2
2
c (1 v c )
(1 v c )
t  0
This magic contraction is now
known as the.........
Einstein - Lorenz contraction,
since :
1) Lorentz was right!
- - it does contract by that amount!
2) MM were wrong - -there is no aether
3) Einstein had a completely
new Paradigm - - - Special Relativiy!